AP Physics 1 Review Sheet

A.P. Physics 1 First Semester Review Sheet Chapter 1: Introduction to Physics ¥ Review types of zeros and the rules for significant digits ¥ Review mass vs. weight, precision vs. accuracy, and dimensional analysis problem solving. Chapter 2: One-Dimensional Kinematics A. Velocity vave = Dx x f - xi = Dt t f - t i Equations for average velocity: ¥ In a position-versus-time graph for constant velocity, the slope of the line gives the average velocity. See Table 1. Instantaneous velocity can be determined from the slope of a line tangent to the curve at a particular point on a position-versus-time graph. ¥ ¥ Use vave = and 1 vave = (v f + vi ) 2 ¥ ΔxTotal to calculate the average velocity for an entire journey if given information about the ΔtTotal various legs of the journey. B. Acceleration aave = D v v f - vi = Dt t f - ti ¥ Equation for average acceleration: ¥ In a velocity-versus-time graph for constant acceleration, the slope of the line gives acceleration and the area under the line gives displacement. See Table 1. Acceleration due to gravity = g = 9.81 m/s2. (Recall a = - g = -9.81 m/s2) ¥ Table 1: Graphing Changes in Position, Velocity, and Acceleration Constant Constant Constant Ball Thrown Position Velocity Acceleration Upward Position Versus Time: Slope = vave x x t Velocity Versus Time: t x t Slope = aave v v t Accelera -tion Versus Time: x v t t Slope = -9.81 m/s2 v t t a = -9.81m/s2 a a t a t a t t A.P. Physics 1 First Semester Review Sheet, Page 2 Table 2: Comparing the Kinematic Equations Kinematic Equations Missing Variable x = xo + vavet a v = vo + at Dx 1 x = xo + v ot + at 2 2 v final v2 = vo2 + 2 aDx Dt Chapter 3: Vectors in Physics A. Vectors ¥ Vectors have both magnitude and direction whereas scalars have magnitude but no direction. ¥ Examples of vectors are position, displacement, velocity, linear acceleration, tangential acceleration, centripetal acceleration, applied force, weight, normal force, frictional force, tension, spring force, momentum, gravitational force, and electrostatic force. ¥ Vectors can be moved parallel to themselves in a diagram. ¥ Vectors can be  added in any order. See Table 3 for vector addition.  ¥ For vector r Δx = r cosθ ¥ at angle q to the x-axis, the x- and y-components for Δy = r sinθ . and The magnitude of vector nearest x-axis is ¥ ¥ ¥  r is r = Δx 2 + Δy 2 r can be calculated from and the direction angle for  r relative to the ⎛ Δy ⎞ θ = tan−1 ⎜ ⎟ . ⎝ Δx ⎠ To subtract a vector, add its opposite. Multiplying or dividing vectors by scalars results in vectors. In addition to adding vectors mathematically as shown in the table, vectors can be added graphically. Vectors can be drawn to scale and moved parallel to their original positions in a diagram so that they are all positioned head-to-tail. The length and direction angle for the resultant can be measured with a ruler and protractor, respectively. B. Relative Motion ¥ Relative motion problems are solved by a special type of vector addition. ¥ ¥ For example, the velocity of object 1 relative to object 3 is given by 2 can be anything.    v13 = v 12 + v 23 Subscripts on a velocity can be reversed by changing the vectorÕs direction:   v12 = −v 21 where object A.P. Physics 1 First Semester Review Sheet, Page 3 Vector Orientation Vectors are parallel: Table 3: Vector Addition Calculational Strategy Used Add or subtract the magnitudes (values) to get the resultant. Determine the direction by inspection. Vectors are perpendicular: Use the Pythagorean Theorem, Dx 2 + Dy 2 = r 2 , to get the resultant, r , where Dx is parallel to the x-axis and Dy is parallel to the y-axis. æ Dy ö ÷ to get the angle, q , made with the x-axis. è Dx ø Use q = tan -1 ç Vectors are neither parallel nor perpendicular: Adding 2 Vectors Limited usefulness (1) Use the law of cosines to determine the resultant: c 2 = a 2 + b 2 - 2ab cos q (2) Use the law of sines to help determine direction: a b c = = sin A sin B sin C Adding 2 or More Vectors (Vector Resolution Method) Used by most physicists (1) Make a diagram. (2) Construct a vector table. (Use vector, x-direction, and y-direction for the column headings.) (3) Resolve vectors using Δx = r cosθ and Δy = r sinθ when needed. (4) Determine the signs. (5) Determine the sum of the vectors for each direction, Dxtotal and Dy total . (6) Use the Pythagorean Thm to get the resultant, r : Dx total2 + Dy total2 = r 2 æ D ytotal ö ÷ to è D xtotal ø get the angle, q . (7) Use q = tan ç 1 A.P. Physics 1 First Semester Review Sheet, Page 4 Chapter 4: Two-Dimensional Kinematics A. Projectile Motion ¥ See Table 4 to better understand how the projectile motion equations can be derived from the kinematic equations. ¥ The kinematic equations involve one-dimensional motion whereas the projectile motion equations involve two-dimensional motion. Two-dimensional motion means there is motion in both the horizontal and vertical directions. ¥ Recall that the equation for horizontal motion (ex. Dx = v xDt ) and the equations for vertical motion (ex. vy , f = - g Dt , Dy = - ¥ 1 g (Dt )2 , v 2y, f = -2 g Dy ) are independent from each other. 2 ¥ Recall that velocity is constant and acceleration is zero in the horizontal direction. ¥ Recall that acceleration is g = 9.81 m/s2 in the vertical direction. When projectiles are launched at an angle, the range of the projectile is often calculated from Δx = (vi cos θ )Δt and its time of flight is often calculate from D y = ( vi sin q)D t ¥ Projectiles follow a parabolic pathway governed by 1 g( D t)2 . 2 æ g ö 2 y = h-ç x 2÷ è 2v o ø Table 4: Relationship Between the Kinematic Equations and Projectile Motion Equations Kinematic Equations Missing Projectile Motion, Projectile Motion, Variable Zero Launch Angle General Launch Angle Assumptions made: Assumptions made: a = - g and vo , y = 0 a = - g , vo , x = vo cos q , and vo , y = vo sin q x = xo + vavet a Dx = vxt Dx = (v o cosq )t where vx = const . where v x = const . v = vo + at Dx v y = - gt vy = vo sin q - gt 1 x = xo + v ot + at 2 2 v final 1 Dy = - gt2 2 1 Dy = ( v o sin q)t - gt 2 2 v2 = vo2 + 2 aDx Dt vy2 = -2 g Dy vy 2 = vo 2 sin2 q - 2gD y  ¥ ¥ ¥ For an object in free fall, the object stops accelerating when the force of air resistance, F Air , equals  the weight, W . The object has reached its maximum velocity, the terminal velocity. When a quarterback throws a football, the angle for a high, lob pass is related to the angle for a low, bullet pass. When both footballs are caught by a receiver standing in the same place, the sum of the launch angles is 90o. In distance contests for projectiles launched by cannons, catapults, trebuchets, and similar devices, projectiles achieve the farthest distance when launched at a 45o angle. A.P. Physics 1 First Semester Review Sheet, Page 5 æ v o2 ö ÷sin 2q è g ø ¥ The range of a projectile launched at initial velocity v o and angle q is R = ç ¥ The maximum height of a projectile above its launch site is y max = v o2 sin2 q 2g Chapter 5: NewtonÕs Laws of Motion Table 5: NewtonÕs Laws of Motion Modern Statement for Law Translation If the net force on an object is An object at rest will remain at NewtonÕs First Law: (Law of Inertia) zero, its velocity is constant. rest. An object in motion will Recall that mass is a measure of remain in motion at constant inertia. velocity unless acted upon by an external force. NewtonÕs Second Law: An object of mass m has an F net = ma  acceleration  a given by the net ∑ F divided by m . That  ∑F is a = m force NewtonÕs Third Law: Recall action-reaction pairs For every force that acts on an object, there is a reaction force acting on a different object that is equal in magnitude and opposite in direction. For every action, there is an equal but opposite reaction. A. Survey of Forces ¥ A force is a push or a pull. The unit of force is the Newton (N); 1 N = 1 kg-m/s2 ¥ See NewtonÕs laws of motion in Table 5. Common forces on a moving object include an applied force, a frictional force, a weight, and a normal force. ¥ Contact forces are action-reaction pairs of forces produced by physical contact of two objects. Review calculations regarding contact forces between two or more boxes. ¥ Field forces like gravitational forces, electrostatic forces, and magnetic forces do not require direct contact. They are studied in later chapters. ¥ Forces on objects are represented in free-body diagrams. They are drawn with the tails of the vectors originating  at an objectÕs center of mass. ¥ Weight, W , is the gravitational force exerted by Earth on an object whereas mass, m , is a measure of the quantity of matter in an object ( W = mg ). Mass does not depend on gravity. ¥ Apparent weight, W a , is the force felt from contact with the floor or a scale in an accelerating system. For example, the sensation of feeling heavier or lighter in an accelerating elevator.    ¥ The normal force, N , is perpendicular to the contact  surface  along which an object moves or is capable of moving. Thus, for an object on a level surface, N and W are equal in size but opposite in direction. However, for an object on a ramp, this statement is not true because N is perpendicular to the surface of the ramp.  ¥ Tension, T , is the force transmitted through a string. The tension is the same throughout the length of an ideal string. A.P. Physics 1 First Semester Review Sheet, Page 6 ¥ The force of an ideal spring stretched or compressed by an amount x is given by HookeÕs Law,  F x = −kx . Note that if we are only interested in magnitude, we use F = kx where k is the spring or force constant. HookeÕs Law is also used for rubber bands, bungee cords, etc. Chapter 6: Applications of NewtonÕs Laws A. Friction ¥ ¥ ¥  FS ,max where F S ,max is the max. force due to static friction. N  FK Coefficient of kinetic friction = µ K = where FK is the force due to kinetic friction. N Coefficient of static friction = µS = A common lab experiment involves finding the angle at which an object just begins to slide down a ramp. In this case, a simple expression can be derived to determine the coefficient of static friction: µS = tan q . Note that this expression is independent of the mass of the object. B. NewtonÕs Second Law Problems (Includes Ramp Problems) 1. Draw a free-body diagram to represent the problem. 2. If the problem involves a ramp, rotate the x- and y-axes so that the x-axis corresponds to the surface of the ramp. 3. Construct a vector table including all of the forces in the free-body diagram. For the vector tableÕs column headings, use vector, x-direction, and y-direction. 4. Determine the column total in each direction: a. If the object moves in that direction, the total is ma . b. If the object does not move in that direction, the total is zero. c. Since this is a NewtonÕs Second Law problem, no other choices besides zero and ma are possible. 5. Write the math equations for the sum of the forces in the x- and y-directions, and solve the problem. It is often helpful to begin with the y-direction since useful expressions are derived that are sometimes helpful later in the problem. Recall that the math equations regarding friction and weight are often substituted into the math equations to help solve the problem. C. Equilibrium  ∑F=0. ¥ An object is in translational equilibrium if the net force acting on it is zero, ¥ Equivalently, an object is in equilibrium if it has zero acceleration. ¥ If a vector table is needed for an object in equilibrium, then ¥ Typical problems involve force calculations for objects pressed against walls and tension calculations for pictures on walls, laundry on a clothesline, hanging baskets, pulley systems, traction systems, connected objects, etc.  ∑F x  = 0 and ∑F y = 0. D. Connected Objects ¥ Connected objects are linked physically, and thus, they are also linked mathematically. For example, objects connected by strings have the same magnitude of acceleration. ¥ When a pulley is involved, the x-y coordinate axes are often rotated around the pulley so that the objects are connected along the x-axis. ¥ A classic example of a connected object is an AtwoodÕs Machine, which consists of two masses connected by a string that passes over a single pulley. The acceleration for this system is given by æ m -m 1 ö a=ç 2 ÷g . è m1 + m2 ø A.P. Physics 1 First Semester Review Sheet, Page 7 Chapter 7: Work and Kinetic Energy A. Work ¥ A force exerted through a distance performs mechanical work. ¥ When force and distance are parallel, W = Fd with Joules (J) or Nm as the unit of work. ¥ When force and distance are at an angle, only the component of force in the direction of motion is used to compute the work: W = ( F cosq )d = Fd cosq o 2 2 ¥ Work is negative if the force opposes the motion (q >90 ). Also, 1 J = 1 Nm = 1 kg-m /s . n ¥ If more than one force does the work, then WTotal = åWi i =1 1 1 WTotal = DK = K f - Ki = mv f2 - mvi2 2 2 ¥ The work-kinetic energy theorem states that ¥ See Table 6 for more information about kinetic energy. ¥ In thermodynamics, æ Aö æ F ö W = Fd = Fd ç ÷ = ç ÷ ( Ad ) = P DV for work done on or by a gas. è Aø è Aø Kinetic Energy Type Table 6: Kinetic Energy Equation Comments Used to represent kinetic energy in most conservation of mechanical energy problems. Kinetic Energy as a Function of Motion: K= Kinetic Energy as a Function of Temperature: 3 æ1 2ö ç mv ÷ = Kave = kT 2 è2 ø ave 1 2 mv 2 Kinetic theory relates the average kinetic energy of the molecules in a gas to the Kelvin temperature of the gas. B. Determining Work from a Plot of Force Versus Position ¥ In a plot of force versus position, work is equal to the area between the force curve and the displacement on the x-axis. For example, work can be easily computed using W = Fd when rectangles are present in the diagram. ¥ For the case of a spring force, the work to stretch or compress a distance x from equilibrium is 1 2 kx . On a plot of force versus position, work is the area of a triangle with base x 2 (displacement) and height kx (magnitude of force using HookeÕs Law, F = kx ). W= C. Determining Work in a Block and Tackle Lab ¥ The experimental work done against gravity, W Load , is the same as the theoretical work done by the spring scale, W Scale . ¥   WOutput = WLoad = Fd Load = W d Load = mgd Load ¥ W Input = W Scale = Fd Scale where F = force read from the spring scale and d Scale = distance the scaled ¥ moved from its original position. Note that the force read from the scale is ½ of the weight when two strings are used for the pulley system, and the force read is ¼ of the weight when four strings are used. where d Load = distance the load is raised. A.P. Physics 1 First Semester Review Sheet, Page 8 D. Power W t ¥ P= ¥ 1 W = 1 J/s P = Fv or and with Watts (W) as the unit of Power. 746 W = 1 hp where hp is the abbreviation for horsepower. Chapter 8: Potential Energy and Conservation of Energy A. Conservative Forces Versus Nonconservative Forces 1. Conservative Forces ¥ A conservative force does zero total work on any closed path. In addition, the work done by a conservative force in going from point A to point B is independent of the path from A to B. In other words, we can use the conservation of mechanical energy principle to solve complex problems because the problems only depend on the initial and final states of the system. ¥ In a conservative system, the total mechanical energy remains constant: Ei = E f . Since E = U + K , it follows that Ui + Ki = U f + K f . See Table 6 for kinetic energy, K , and Table 7 for potential energy, U , for additional information. ¥ For a ball thrown upwards, describe the shape of the kinetic energy, potential energy, and total energy curves on a plot of energy versus time. ¥ Examples of conservative forces are gravity and springs. 2. Nonconservative Forces ¥ The work done by a nonconservative force on a closed path is not zero. In addition, the work depends on the path going from point A to point B. ¥ In a nonconservative system, the total mechanical energy is not constant. The work done by a nonconservative force is equal to the change in the mechanical energy of a system: WNonconservative = Wnc = DE = E f - Ei . ¥ Examples of nonconservative forces include friction, air resistance, tension in ropes and cables, and forces exerted by muscles and motors. Potential Energy Type Table 7: Potential Energy Equation Gravitational Potential Energy: U = mgh Gravitational Potential Energy Between Two Point Masses: U = -G m1m2 r Good approximation for an object near sea level on the EarthÕs surface. where G = 6.67 x 10-11 Nm2/kg2 = Universal Gravitation Constant Elastic Potential Energy: Comments Works well at any altitude or distance between objects in the universe; recall that r is the distance between the centers of the objects. Useful for springs, rubber bands, 1 2 kx where k is the force bungee cords, and other 2 stretchable materials. (spring) constant and x is the U= distance the spring is stretched or compressed from equilibrium. A.P. Physics 1 First Semester Review Sheet, Page 9 Chapter 9: Linear Momentum and Collisions A. Momentum ¥   p = mv with kg-m/s as the unit of momentum. n   pTotal = ∑ p i . In a system having several objects, ¥ NewtonÕs second law can be expressed in terms of momentum. The net force acting on an object is  ¥ Linear momentum is given by i=1  Δp F net = Δt              − p − mvi m(v f − vi ) mΔ v p mv Δp i f f When mass is constant, = F net = = ma . = = = Δt Δt Δt Δt Δt     I = F ave Δt = Δ p The impulse-momentum theorem states that where the quantity F ave Δt is called the impulse, I . For problem-solving purposes, a more useful form of the impulse-momentum theorem is FΔt = m(v f − vi ) . !" !" A practical application of the impulse-momentum theorem, F ave Δt = Δ p , involves auto air bags. In an automobile collision, the change in momentum, Δp , remains constant. Thus, an increase in  collision time, Dt , will result in a decreased force of impact, F , reducing personal injury. equal to the rate of change in its momentum: ¥ ¥ ¥ B. Collisions ¥ For a system of objects, the conservation of linear momentum principle states that the net momentum is conserved if the net external force acting on the system is zero. In other words, pi = p f . ¥ For an elastic collision, m1v1i + m2v2i = m1v1 f + m2v 2 f and for a perfectly inelastic collision, m1 v1 i + m2 v2 i = (m1 + m2 )v f . See the collision types in Table 8. ¥ In an elastic collision in one dimension where mass m1 is moving with an initial velocity vo , and mass m2 is initially at rest, the velocities of the masses after the collision are: æ m - m2 ö v1, f = ç 1 ÷ vo è m1 + m2 ø and æ 2m1 ö v2, f = ç ÷ vo . è m1 + m2 ø Collision Type and Example Elastic: No permanent deformation occurs; billiard balls collide. Inelastic: Permanent deformation occurs; Most automobile collisions. Perfectly Inelastic: Permanent deformation occurs and objects lock together moving as a single unit; train cars collide and lock together. Table 8: Collision Types Do the Objects Is Momentum Stick Together? Conserved? Is Kinetic Energy Conserved? No Yes Yes No Yes No Yes Yes No A.P. Physics 1 First Semester Review Sheet, Page 10 Chapter 10: Rotational Kinematics and Energy A. Rotational Motion Ds where Ds is arc length and r is radius. r ¥ Angular position, Dq , in radians is given by Dq = ¥ Recall that θ (rad) = ⎜ ¥ ¥ Counterclockwise (CCW) rotations are positive, and clockwise (CW) rotations are negative. In rotational motion, there are two types of speeds (angular speed and tangential speed) and three types of accelerations (angular acceleration, tangential acceleration, and centripetal acceleration). See Table 9 for a comparison. Since velocity is a vector, there are two ways that an acceleration can be produced: (1) changing the velocityÕs magnitude and (2) changing the velocityÕs direction. In centripetal acceleration, the velocityÕs direction changes. When a person drives a car in a circle at constant speed, the car has a centripetal acceleration due to its changing direction, but it has no tangential acceleration due to its constant speed. The total acceleration of a rotating object is the vector sum of its tangential and centripetal accelerations. ¥ ¥ ¥ ⎛ 2π radians ⎞ ⎟⎠ θ (deg) . ⎝ 360o Table 9: Comparing Angular and Tangential Speed and Angular, Tangential, and Centripetal Acceleration Calculation Equations Units, Comments Angular Speed: Tangential Speed: Angular Acceleration: Tangential Acceleration: wave = Dq Dt v t = rw a ave = Dw Dt at = ra ¥ ¥ radians/s Same value for horses A and B, side-by-side on a merrygo-round. ¥ ¥ m/s Different values for horses A and B, side-by-side on a merry-go-round. ¥ radians/s2 Same value for horses A and B, side-by-side on a merrygo-round. ¥ ¥ ¥ Centripetal Acceleration: ac = vt 2 = r w2 r ¥ ¥ m/s2 Different values for horses A and B, side-by-side on a merry-go-round. m/s2 ac is perpendicular to at with ac directed toward the center of the circle and at tangent to it. A.P. Physics 1 First Semester Review Sheet, Page 11 mv2t = mr w 2 r ¥ Centripetal force, FC , is a force that maintains circular motion: F C = ma C = ¥ The period, T , is the time required to complete one full rotation. If the angular speed is constant, then T= ¥ 2p w . The equations for rotational kinematics are the same as the equations for linear kinematics. See Table 10 for a comparison. Table 10: Kinematic equations for Rotational Motion Linear Equations Angular Equations ¥ ¥ ¥ ¥ x = xo + vavet θ = θo + ω avet v = vo + at ω = ω o + αt 1 x = xo + vot + at 2 2 1 θ = θo + ω ot + α t 2 2 v2 = vo2 + 2 aDx ω 2 = ω o 2 + 2α (Δθ ) A comparison of linear and angular inertia, velocity, acceleration, NewtonÕs second law, work, kinetic energy, and momentum are presented in Table 11. The moment of inertia, I , is the rotational analog to mass in linear systems. It depends on the shape or mass distribution of the object. In particular, an object with a large moment of inertia is difficult to start rotating and difficult to stop rotating. See Table 10-1 on p.298 for moments of inertia for uniform, rigid objects of various shapes and total mass. The greater the moment of inertia, the greater an objectÕs rotational kinetic energy. An object of radius r , rolling without slipping, translates with linear speed v and rotates with angular speed v r w= . A.P. Physics 1 First Semester Review Sheet, Page 12 Table 11: Comparing Equations for Linear Motion and Rotational Motion Measurement or Calculation Linear Equations Angular Equations Inertia: Average Velocity: Average Acceleration: I = kmr 2 where k = a constant Mass, m v ave = Dx Dt wave = Dq Dt aave = Dv Dt a ave = Dw Dt NewtonÕs Second Law: F net = ma t = Ia = F^r Work: W = Fd cos q W = tq Kinetic Energy: Momentum: K= 1 2 mv 2 K= p = mv 1 2 Iw 2 L = Iw Chapter 11: Rotational Dynamics and Static Equilibrium A. Torque ¥ A tangential force, F , applied at a distance, r , from the axis of rotation produces a torque in Nm. (Since F is perpendicular to r , this is sometimes written as t = F^r .) ¥ ¥ ¥ ¥ t = rF A force applied at an angle to the radial direction produces the torque t = rF sinq Counterclockwise torques are positive, and clockwise torques are negative. The rotational analog of force, F = ma , is torque, t = Ia , where I = moment of inertia and a = angular acceleration. The conditions for an object to be in static equilibrium are that the total force and the total torque acting on the object must be zero: Fx = 0, Fy = 0, t = 0. Related problems often involve å å å bridges, scaffolds, signs, and rods held by wires. B. Solving Static Equilibrium Problems 1. Construct a diagram showing all of the forces. 2. Create an equation adding the forces together. ¥ Remember to enter correct signs in your force equation. For example, upward forces are positive and downward forces are negative. ¥ Since the object is not moving, set the force equation equal to zero. ( ∑F = 0) A.P. Physics 1 First Semester Review Sheet, Page 13 3. Create an equation adding the torques together. ¥ If you are not sure what to do: ¥ Start by writing the forces again leaving some space between them. ¥ Multiply each force by an appropriate distance. ¥ Each force and distance pair will have the same subscript. ¥ Choose an axis of rotation in your diagram. It is helpful if your choice eliminates one of the two unknown forces. Then you can solve for the other force. ¥ Remember to enter correct signs in your torque equation. Consider whether each force will create a counterclockwise or clockwise rotation resulting in a positive or negative torque, respectively. ¥ Since the object is not moving, set the torque equation equal to zero. ( åt = 0) 4. Substitute numbers in your torque equation and solve for the unknown force. 5. Substitute the force you determined in the last step into the force equation and solve for the other unknown force. C. Angular Momentum ¥ The rotational analog of momentum, p = mv , is angular momentum, L = I w in kg-m2/s, where I = moment of inertia and w = angular velocity. ¥ If the net external torque acting on a system is zero, its angular momentum is conserved and Li = L f . D. Simple Machines ¥ All machines are combinations or modifications of six fundamental types of machines called simple machines. ¥ Simple machines include the lever, inclined plane, wheel and axle, wedge, pulley(s), and screw. MA = Fout din . = Fin dout ¥ Mechanical advantage, MA , is defined as ¥ much force or distance is multiplied by a machine. Efficiency is a measure of how well a machine works, and % Efficiency is calculated using ⎛W ⎞ % Efficiency = ⎜ out ⎟ 100 ⎝ Win ⎠ ( ) It is a number describing how where Wout is the work output and Win is the work input. Chapter 12: Gravity A. General Concepts About Gravity and KeplerÕs Laws ¥ NewtonÕs Law of Universal Gravitation shows that the force of gravity between two point masses, m1 and m2 , separated by a distance r is -11 ¥ m1m2 r2 where G is the universal gravitation constant, 2 G = 6.67x10 Nm /kg . Remember that r is the distance between the centers of the point masses. In NewtonÕs Law of Universal Gravitation, notice that the force of gravity decreases with distance, r , as ¥ 2 F=G 1 . This is referred to as an inverse square dependence. r2 The superposition principle can be applied to gravitational force. If more than one mass exerts a gravitational force on a given object, the net force is simply the vector sum of each individual force. (The superposition principle is also used for electrostatic forces, electric fields, electric potentials, electric potential energies, and wave interference.) A.P. Physics 1 First Semester Review Sheet, Page 14 ¥ Replacing the Earth with a point mass at its center, the acceleration due to gravity at the surface of the Earth is given by g= GM Earth . 2 REarth A similar equation is used to calculate the acceleration due to gravity at the surface of other planets and moons in the Solar System. M Earth ¥ At some altitude, h , above the Earth, ¥ acceleration due to gravity. In 1798, Henry Cavendish first determined the value of G, which allowed him to calculate the mass of the Earth using ¥ ¥ M Earth = gh = G 2 gREarth . G Using Tycho BraheÕs observations concerning the planets, Johannes Kepler formulated three laws for orbital motion as shown in Table 12. Newton later showed that each of KeplerÕs laws follows as a direct consequence of the universal law of gravitation. As previously mentioned in the potential energy table, the gravitational potential energy, U , between two point masses m1 and m2 separated by a distance r is ¥ can be used to calculate the ( R Earth + h) 2 U = -G m1m2 . r This equation is used in mechanical energy conservation problems for astronomical situations. Energy conservation considerations allow the escape speed to be calculated for an object launched from the surface of the Earth: Law 1st Law: ve = 2GM Earth = 11,200 m/s » 25,000 mi/h. R Earth Table 12: KeplerÕs Laws of Orbital Motion Modern Statement for Law Alternate Description Planets follow elliptical orbits, with the The paths of the planets are ellipses, with Sun at one focus of the ellipse. the center of the Sun at one focus. 2nd Law: As a planet moves in its orbit, it sweeps out an equal amount of area in an equal amount of time. An imaginary line from the Sun to a planet sweeps out equal areas in equal time intervals. 3rd Law: The period, T , of a planet increases as its mean distance from the Sun, r , raised to the 3/2 power. That is, The square of a planetÕs period, T 2 , is 3 proportional to the cube of its radius, r . æ 2p ö 3/ 2 T =ç ÷ r = (constant) r3/ 2 ç GM ÷ S ø è That is, T 2 = 4p 2 3 r = (constant) r 3 GM S where M S = mass of the Sun.